Problem: Is ${36226}$ divisible by $9$ ?
Answer: A number is divisible by $9$ if the sum of its digits is divisible by $9$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {36226}= &&{3}\cdot10000+ \\&&{6}\cdot1000+ \\&&{2}\cdot100+ \\&&{2}\cdot10+ \\&&{6}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {36226}= &&{3}(9999+1)+ \\&&{6}(999+1)+ \\&&{2}(99+1)+ \\&&{2}(9+1)+ \\&&{6} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {36226}= &&\gray{3\cdot9999}+ \\&&\gray{6\cdot999}+ \\&&\gray{2\cdot99}+ \\&&\gray{2\cdot9}+ \\&& {3}+{6}+{2}+{2}+{6} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $9$ , so the first four terms must all be multiples of $9$ That means that to figure out whether the original number is divisible by $9 $ , all we need to do is add up the digits and see if the sum is divisible by $9$ . In other words, ${36226}$ is divisible by $9$ if ${ 3}+{6}+{2}+{2}+{6}$ is divisible by $9$ Add the digits of ${36226}$ $ {3}+{6}+{2}+{2}+{6} = {19} $ If ${19}$ is divisible by $9$ , then ${36226}$ must also be divisible by $9$ ${19}$ is not divisible by $9$, therefore ${36226}$ must not be divisible by $9$.